/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 57 An electrical conductor designed... [FREE SOLUTION] | 91Ó°ÊÓ

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An electrical conductor designed to carry large currents has a circular cross section 2.50 \(\mathrm{mm}\) in diameter and is 14.0 \(\mathrm{m}\) long. The resistance between its ends is \(0.104 \Omega .\) (a) What is the resistivity of the material? (b) If the electric-field magnitude in the conductor is 1.28 \(\mathrm{V} / \mathrm{m}\) , what is the total current? (c) If the material has \(8.5 \times 10^{28}\) free electrons per cubic meter, find the average drift speed under the conditions of part (b).

Short Answer

Expert verified
(a) \(3.64 \times 10^{-8} \Omega \cdot \mathrm{m}\); (b) 172.31 A; (c) \(2.63 \times 10^{-4} \mathrm{m/s}\).

Step by step solution

01

Calculate the cross-sectional area

The resistance of a conductor is related to its resistivity by the formula \( R = \rho \frac{L}{A} \), where \( R \) is the resistance, \( \rho \) is the resistivity, \( L \) is the length, and \( A \) is the cross-sectional area. The area of a circle is given by \( A = \pi r^2 \), where \( r \) is the radius. The diameter is 2.50 mm, so the radius \( r = \frac{2.50 \, \mathrm{mm}}{2} = 1.25 \, \mathrm{mm} = 1.25 \times 10^{-3} \, \mathrm{m} \). Substitute \( r \) into the area formula to find \( A \):\[ A = \pi (1.25 \times 10^{-3} \, \mathrm{m})^2 = \pi \times 1.5625 \times 10^{-6} \, \mathrm{m}^2 = 4.91 \times 10^{-6} \, \mathrm{m}^2 \]
02

Calculate resistivity from resistance

Using the formula \( R = \rho \frac{L}{A} \), we can solve for the resistivity \( \rho \):\[ \rho = R \frac{A}{L} \]Given \( R = 0.104 \, \Omega \), \( A = 4.91 \times 10^{-6} \, \mathrm{m}^2 \), and \( L = 14.0 \, \mathrm{m} \), substitute these values to find \( \rho \):\[ \rho = 0.104 \, \Omega \times \frac{4.91 \times 10^{-6} \, \mathrm{m}^2}{14.0 \, \mathrm{m}} = 3.64 \times 10^{-8} \, \Omega \cdot \mathrm{m} \]
03

Calculate the total current from electric field

The relationship between the electric field \( E \), current \( I \), resistance \( R \), and voltage \( V \) is given by Ohm's Law: \( V = IR \). Also, \( V = EL \), hence \( IR = EL \). Reorganizing gives \( I = \frac{EL}{R} \).Given \( E = 1.28 \, \mathrm{V/m} \), \( L = 14.0 \, \mathrm{m} \), and \( R = 0.104 \, \Omega \), substitute to find \( I \):\[ I = \frac{1.28 \, \mathrm{V/m} \times 14.0 \, \mathrm{m}}{0.104 \, \Omega} = 172.31 \, \mathrm{A} \]
04

Calculate average drift speed

The current \( I \) is related to the number density of charge carriers \( n \), the charge of an electron \( e \), the cross-sectional area \( A \), and the drift velocity \( v_d \) by the formula \( I = nAve \).Reorganize the formula to find the drift velocity \( v_d \):\[ v_d = \frac{I}{nAe} \]Given \( I = 172.31 \, \mathrm{A} \), \( n = 8.5 \times 10^{28} \, \mathrm{m}^{-3} \), \( A = 4.91 \times 10^{-6} \, \mathrm{m}^2 \), and \( e = 1.60 \times 10^{-19} \, \mathrm{C} \), substitute to find \( v_d \):\[ v_d = \frac{172.31 \, \mathrm{A}}{8.5 \times 10^{28} \, \mathrm{m}^{-3} \cdot 4.91 \times 10^{-6} \, \mathrm{m}^2 \cdot 1.60 \times 10^{-19} \, \mathrm{C}} = 2.63 \times 10^{-4} \, \mathrm{m/s} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electric Field
The electric field in a conductor fundamentally affects how electricity flows. You can think of it as a force that pushes the electrical charges along the conductor, in this case, the wire. This force is measured in volts per meter (V/m). When you have a stronger electric field, it means the charges are pushed harder, allowing them to move faster through the conductor.
In practical terms, if you imagine water guiding in a pipe, the electric field is like the water pressure. A higher pressure makes the water flow faster, just like a higher electric field makes the electrons move faster.
Additionally, the electric field can be linked with voltage through the length of the conductor. In the formula, when we write \(V = EL\), it highlights that the voltage (V) across the length is the result of multiplying the electric field (E) with the length (L) of the path the charges travel along. This relationship is crucial as it helps determine how much energy is used to move charges from one point to another in the conductor.
Drift Velocity
Drift velocity is an essential concept when discussing how current travels through a conductor. It describes the average speed at which free electrons move through a material when subjected to an electric field. Although electric currents move very quickly to light a bulb or power a gadget, the individual electron moves surprisingly slowly.
In our example, the drift velocity can be calculated using the formula \(v_d = \frac{I}{nAe}\), where \(I\) is the current, \(n\) is the number of charge carriers per cubic meter, \(A\) is the cross-sectional area, and \(e\) is the charge of an electron.
Interestingly, even though drift speed might seem small, approximately \(2.63 \times 10^{-4} \ \mathrm{m/s}\) in our problem, the effect is instantaneous on the large scale because of the energy transfer in the conductor.
This concept explains why turning on a switch causes a light to turn on almost immediately, even though each electron takes longer to travel across the wire.
Cross-sectional Area
The cross-sectional area of a conductor like a wire significantly impacts its electrical properties, like resistance and drift velocity. In simple terms, it's the size of the surface area if you cut through the wire and look at that circle face-on.
The formula for calculating the area of a circle is \(A = \pi r^2\). Therefore, we need the radius, which is half the diameter, to determine this area. For a wire with a diameter of 2.50 mm, the radius is 1.25 mm or \(1.25 \times 10^{-3} \ \mathrm{m}\). This makes the cross-sectional area about \(4.91 \times 10^{-6} \ \mathrm{m}^2\).
The area affects how much space is available for electrons to travel through. A larger cross-sectional area reduces resistance because it gives more room for the electrons, just like a wider pipe lets more water through with less resistance. As a result, a wire with a greater area can carry more current without overheating.

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Most popular questions from this chapter

The potential difference between points in a wire 75.0 \(\mathrm{cm}\) apart is 0.938 \(\mathrm{V}\) when the current density is \(4.40 \times 10^{7} \mathrm{A} / \mathrm{m}^{2}\) . What are (a) the magnitude of \(\overrightarrow{\boldsymbol{E}}\) in the wire and \((\mathrm{b})\) the resistivity of the material of which the wire is made?

A copper wire has a square cross section 2.3 \(\mathrm{mm}\) on a side. The wire is 4.0 \(\mathrm{m}\) long and carries a current of 3.6 \(\mathrm{A}\) . The density of free electrons is \(8.5 \times 10^{28} / \mathrm{m}^{3}\) . Find the magnitudes of (a) the current density in the wire and \((b)\) the electric field in the wire. (c) How much time is required for an electron to travel the length of the wire?

Light Bulbs. The power rating of a light bulb (such as a \(100-W\) bulb) is the power it dissipates when connected across a \(120-V\) potential difference. What is the resistance of (a) a \(100-W\) bulb and (b) a \(60-\mathrm{W}\) bulb? (c) How much current does each bulb draw in normal use?

A battery-powered global positioning system (GPS) receiver operating on 9.0 \(\mathrm{V}\) Uraws a current of 0.13 \(\mathrm{A}\) . How much electrical energy does it consume during 1.5 \(\mathrm{h} ?\)

A \(5.00-\mathrm{A}\) current runs through a 12 -gauge copper wire (diameter 2.05 \(\mathrm{mm} )\) and through a light bulb. Copper has \(8.5 \times 10^{28}\) free electrons per cubic meter. (a) How many electrons pass through the light bulb each second? (b) What is the current density in the wire? (c) At what speed does a typical electron pass by any given point in the wire? (d) If you were to use wire of twice the diameter, which of the above answers would change? Would they increase or decrease?

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